# How to prove $|a+b|^k \leq 2^{k-1} (|a|^k+|b|^k)$?

Assume that you have any two real numbers $a$ and $b$, and $1\leq k <\infty$, $k \in \mathbb{R}$.

How would you prove the inequality $|a+b|^k \leq 2^{k-1} (|a|^k+|b|^k)$?

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What about $k$? Is it a positive integer? A positive real? –  Patrick Da Silva Oct 17 '12 at 7:16
Edited, $1 \leq k < \infty$. –  Rojas Azules Oct 17 '12 at 7:17
Sorry, $k \in \mathbb{R}$. –  Rojas Azules Oct 17 '12 at 7:20

Method $1$:
The function $f(x) = x^k$ for $x\geq 0$ is a convex function for $k \geq 1$. Now apply Jensen's inequality.
Method $2$:
Let $\lvert a \rvert \geq \lvert b \rvert$. Let $t = \dfrac{b}a \implies 0 \leq \vert t \vert \leq 1$. We then want to prove that $$\vert 1 + t \vert^k \leq 2^{k-1} \left(1+\vert t \vert^k \right)$$ $$\left \vert \dfrac{1+t}2 \right \vert^k \leq \dfrac{1+\left \vert t \right \vert^k}2$$ $$\left \vert \dfrac{1+t}2 \right \vert^k \leq \left(\dfrac{1+ \vert t \vert}{2} \right)^k$$ Setting $y = \vert t \vert \in [0,1]$, we want to prove that $$\left(\dfrac{1+ y}{2} \right)^k \leq \dfrac{1+y^k}2$$ $$f(y) = \dfrac{1+y^k}2 - \left(\dfrac{1+ y}{2} \right)^k$$ $$f'(y) = \dfrac{ky^{k-1}}2 - \dfrac{k}{2^k}(1+y)^{k-1} = \dfrac{k}2 \left(y^{k-1} - \left(\dfrac{1+y}2 \right)^{k-1} \right)$$ For $y \in [0,1]$, $y \leq \dfrac{1+y}2$, and hence $y^k \leq \left( \dfrac{1+y}2 \right)^k$. This means that $f'(y) < 0$ and hence $f(y)$ is decreasing. Hence, $$f(y) > f(1)$$ which implies $$\left(\dfrac{1+ y}{2} \right)^k \leq \dfrac{1+y^k}2$$